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Electronic Journal of Qualitative Theory of Differential Equations
2008, No. 11, 1-12; http://www.math.u-szeged.hu/ejqtde/

ON THE OSCILLATION OF SECOND ORDER
NONLINEAR NEUTRAL DELAY DIFFERENCE
EQUATIONS
A.K.Tripathy
Department of Mathematics
Kakatiya Institute of Technology and Science
Warangal-506015, INDIA
e-Mail: arun tripathy70@rediffmail.com
Abstract
In this paper sufficient conditions are obtained for oscillation of all solutions of a
class of nonlinear neutral delay difference equations of the form
∆2 (y(n) + p(n)y(n − m)) + q(n)G(y(n − k)) = 0
under various ranges of p(n). The nonlinear function G, G ∈ C(R, R) is either sublinear or superlinear.
Mathematics Subject classification (2000): 39 A 10, 39 A 12
Keywords : Difference equations; Oscillation; Nonlinear.

1

Introduction

Recently, a good deal of work has been published on the oscillation theory of difference equations. Most of the works in first and higher order neutral delay difference equations
are concerned with the study of the behaviour of the solution which oscillates or tends to zero
(see [3], [4], [5], [7]). But very few papers are available on oscillatory higher order nonlinear
delay difference equations.

EJQTDE, 2008 No. 11, p. 1

In [7], authors Parhi and Tripathy has considered a class of nonlinear neutral delay
difference equations of higher order of the form
∆m [y(n) + p(n)y(n − m)] + q(n)G(y(n − k)) = 0,

(E)

where m ≥ 2. They have obtained the results which hold good when G is sub linear only.
However, the behaviour of solutions of (E) under the superlinear nature of G is still in
progress. In fact, various ranges of p(n) are restricting for all solutions as oscillatory.
In this paper, author has studied the second order nonlinear neutral delay difference
equation of the form

∆2 [y(n) + p(n)y(n − m)] + q(n)G(y(n − k)) = 0, n ≥ 0,

(1)

where ∆ is the forward difference operator defined by ∆ y(n) = y(n + 1) − y(n), p, q are
real valued functions defined on N(0) = {0, 1, 2, ...} such that q(n) ≥ 0, G ∈ C(R, R) is
nondecreasing and x G(x) > 0 for x 6= 0 and m > 0, k ≥ 0 are integers. Here, an attempt
is made to establish sufficient conditions under which every solution of Eq.(1) oscillates.
The motivation of present work has come under two directions. Firstly, due to the
work in [6] and second is due to the work in [7], where G is almost sublinear. It is interesting
to observe that unlike differential equation, Eq. (1) is converting immediately into a first
order difference inequality and hence study of both are interrelated hypothetically. In this
regard the work in [8] provides a good input for the completion of the present work.
By a solution of Eq. (1) we mean a real valued function y(n) defined on N(−r) =
{−r, −r + 1, · · ·} which satisfies (1) for n ≥ 0, where r = max{k, m}. If
y(n) = An , n = −r, −r + 1, · · · , 0

(2)

are given, then (1) admits a unique solution satisfying the initial condition (2). A solution

y(n) of (1) is said to be oscillatory, if for every integer N > 0, there exists an n ≥ N such
that y(n)y(n + 1) ≤ 0 : otherwise, it is called nonoscillatory.
The following two results are useful for our discussion in the next sections.
Theorem 1.1 [2]. If q(n) ≥ 0 for n ≥ 0 and
lim inf
n→∞

n−1
X

s=n−k

q(s) > (

k k+1
) ,
k+1

then ∆ x(n) + q(n) x(n − k) ≤ 0, n ≥ 0 can not have an eventually positive solution.
EJQTDE, 2008 No. 11, p. 2

Theorem 1.2 [8]. Assume that lim inf

n−1
P

n→∞ s=n−k

q(s) > 0.

(H) Suppose there are a function g(u) ∈ C(R, R+ ) and a number ǫ > 0 such that
1.

g(u) is nondecreasing in R+ ,

2.

g(−u) = g(u) and lim g(u) = 0,
n→0

∞
R

3.

g(e−u )du < ∞,

0


 G(u)
 u

4.
If
∞
X

n=0



n+k
X






− 1 ≤ g(u), 0 < |u| < ǫ.

q(s)ℓn

s=n

n+k
X
s=n

!

q(s) −

n+k
X

s=n+1



q(s)ℓn 

n+k
X

s=n+1



q(s) = ∞,

then every solution of ∆ x(n) + q(n) G(x(n − k)) = 0 oscillates.
Remark

Corollary 1.3

lim
inf
n→∞

n−1
P

q(s) > 0 implies that

s=n−k

∞
P

q(n) = ∞.

n=0

If all the conditions of Theorem 1.2 are satisfied, then
∆x(n) + q(n)G(x(n − k)) ≤ 0, n ≥ 0

doesn’t possess any eventually positive solution.
The proof follows from the Theorem 1.2.

EJQTDE, 2008 No. 11, p. 3

2

Sublinear Oscillation

This section deals with the sufficient conditions for the oscillation of all solutions of Eq.(1)
when G is sublinear. The following conditions are needed for our use in the sequel.
(H1 )

(H2 )

∞
P

q(n) = ∞

n=0
∞
P

n=m

Q(n) = ∞,

where Q(n) = min{q(n), q(n − m)}
(H3 )

G(u)G(v) ≥ G(uv) for u > 0, v > 0

(H4 )

G(−u) = −G(u), u ∈ R

(H5 )

There exists λ > 0 such that G(u) + G(v) ≥ λ G(u + v), u ∈ R, v ∈ R

(H6 )

±c
R
0

du
G(u)

(H7 )

lim inf

(H8 )

lim
inf
n→∞

|x|→0

Theorem 2.1
oscillatory.

< ∞, c > 0
G(x)
x

≥γ>0

n−1
P

s=n+m−k

k−m k−m+1
q(s) > γb ( k−m+1
)
, b > 0 and k > m.

Let 0 ≤ p(n) ≤ a < ∞. Suppose that (H2 )-(H6 ) hold. Then Eq.(1) is

Proof
Suppose for contrary that y(n) is a nonoscillatory solution of (1). Then there
exists n1 > 0 such that y(n) > 0 or < 0 for n ≥ n1 . Let the former hold. Setting
z(n) = y(n) + p(n)y(n − m)

(3)

∆2 z(n) + q(n)G(y(n − k)) = 0,

(4)

we have from (1)

that is,
∆2 z(n) = −q(n)G(y(n − k)) ≤ 0,
for n ≥ n2 > n1 + r. Hence ∆z(n) is non-increasing. If ∆z(n) < 0, then Eq.(1) becomes
∆z(n + 1) + q(n)G(y(n − k)) = ∆z(n) < 0, n ≥ n2 .

(5)

EJQTDE, 2008 No. 11, p. 4

Using (5) we get
n ≥ n∗2

∆z(n − m + 1) + q(n − m) G (y(n − k − m)) < 0,
and hence for n3 > max{n2 , n∗2 },

0 > ∆z(n + 1) + q(n) G(y(n − k)) + G(a) ∆z(n − m + 1)
+G(a)q(n − m)G(y(n − k − m)).

(6)

Using (H2 ), (H3 ) and (H5 ), the last inequality implies
0 > ∆z(n + 1) + G(a)∆z(n − m + 1) + λQ(n)G(z(n − k)),
where 0 < z(n) ≤ y(n) + a y(n − m). Thus

λQ(n) +

∆z(n − m + 1)
∆z(n + 1)
+ G(a)
< 0,
G(z(n − k))
G(z(n − k))

that is,
λQ(n) +

z(n+2)
Z

z(n+1)

du
+ G(a)
G(u)

z(n−m+2)
Z

z(n−m+1)

dv
< 0,
G(v)

where z(n + 2) < u < z(n + 1), z(n − m + 2) < v < z(n − m + 1) and n − k < n + 1. Hence
for n ≥ n3 ,
λ

n
X

Q(s) +

s=n3

n
X

s=n3

z(s+2)
Z

z(s+1)

n
X
du
+ G(a)
G(u)
s=n3

z(s−m+2)
Z

z(s−m+1)

dv
< 0,
G(v)

that is,
λ

n
X

s=n3

Q(s) +

z(n+2)
Z

z(n3 +1)

du
+ G(a)
G(u)

z(n3Z
−m+2)

z(n3 −m+1)

dv
< 0.
G(v)

Since lim z(n) exists, then the above inequality implies that
n→∞

∞
X

Q(s) < ∞,

s=n3

a contradiction to (H2 ). If ∆z(n) > 0 for n ≥ n2 , then z(n) is nondecreasing and hence there
exists a constant α > 0 such that z(n) > α, n ≥ n∗ . Application of Eq.(4) gives
∆2 z(n) + G(a)∆2 z(n − m) + λQ(n)G(z(n − k)) ≤ 0.
EJQTDE, 2008 No. 11, p. 5

Consequently, for n ≥ n3 > max{n2 , n∗ },
λG(a)Q(n) < −∆2 z(n) − G(a)∆2 z(n − m).
Thus,
∞
X

Q(n) < ∞,

n=n3

a contradiction to (H2 ).
Suppose the later holds. Then setting x(n) = −y(n) > 0, for n ≥ n1 and using (H4 ),
Eq.(1) can be written as
∆2 (x(n) + p(n)x(n − m)) + q(n)G(x(n − k)) = 0.

(7)

Following the above procedure to Eq.(7), similar contradictions can be obtained. Hence the
proof of the theorem is complete.
Remark

The prototype of G satisfying (H3 ), (H4 ) and (H5 ) is




G(u) = α + β |u|λ |u|µ sqn u,
where α ≥ 1, β ≥ 1, λ ≥ 0 and µ ≥ 0.
Example

Consider
1

∆2 [y(n) + (3 + (−1)n )y(n − 1)] + 8 y 3 (n − 2) = 0, n ≥ 0.
Clearly, the above equation satisfies all the conditions of Theorem 2.1 and hence it is oscillatory. In particular, y(n) = (−1)3n is such an oscillatory solution.
Theorem 2.2
Let −1 < b ≤ p(n) ≤ 0. If (H1 ), (H4 ), (H6 ) and m < k hold, then
every solution of (1) oscillates.
Proof
Proceeding as in Theorem 2.1, we get the inequality (5), where ∆z(n) < 0
and z(n) > 0 for n ≥ n2 . Since z(n) ≤ y(n), then (5) becomes
∆z(n + 1) + q(n)G(z(n − k)) < 0, n ≥ n2 .
Following the similar steps of Theorem 2.1, we have a contradiction to (H1 ). If z(n) < 0,
for n ≥ n2 , then y(n) < y(n − m), that is, y(n) is bounded. Consequently, z(n) is bounded,
a contradiction to the fact that ∆z(n) < 0 and z(n) < 0 for n ≥ n2 . Hence ∆z(n) > 0 for
n ≥ n2 . If z(n) > 0, then using the similar argument as in Theorem 2.1, the contradiction is
EJQTDE, 2008 No. 11, p. 6

obtained to (H1 ). Suppose that z(n) < 0, for n ≥ n2 . Then y(n − k) > (1/b)z(n + m − k).
Thus Eq.(1) becomes
1
∆ z(n) + q(n)G z(n) < 0
b


2



due to increasing z(n). Consequently, the last inequality can be made as
1
z(n + 1) < 0
−∆z(n) + q(n)G
b




that is,
1
∆x(n) − q(n)G(x(n + 1) < 0,
b
where x(n) = 1b z(n) > 0. Hence
x(n+1)
Z
du
∆x(n)
1
=−
− q(n) < −
b
G(x(n + 1))
G(x(n + 1))
x(n)

For x(n + 1) < u < x(n), it is immediate to get
N
N
X
1 X
q(n) < −
b n=n2
n=n2

x(n+1)
Z
x(n)

du
=−
G(u)

x(N
Z +1)
x(n2 )

du
G(u)

that is,
∞
1 X
q(n) < − lim
−
N →∞
b n=n2

x(N
Z +1)
x(n2 )

du
< ∞,
G(u)

a contradiction.
The case y(n) < 0 for n ≥ n1 is similar. This completes the proof of the theorem.
Theorem 2.3
Let −∞ < −b ≤ p(n) < −1, b > 0. If (H1 ), (H4 ), (H7 ) and (H8 )
hold, then every bounded solution of (1) oscillates.
Proof
Let y(n) be a bounded nonoscillatory solution of (1). Then there exists n1 > 0
such that y(n) > 0 for n ≥ n1 . The case y(n) < 0 for n ≥ n1 can similarly be dealt with.
Setting z(n) as in (3), we get (4). Hence ∆2 z(n) ≤ 0, for n ≤ n2 > n1 + r implies that ∆z(n)
in nonincreasing. Let ∆z(n) < 0. Then z(n) > 0 or < 0, for n ≥ n2 . Suppose the former
holds. Then lim z(n) exists. On the other hand, ∆2 z(n) < 0 and ∆z(n) < 0 implies that
n→∞
there exists L > 0 such that ∆z(n) < −L, for n ≥ n∗ that is, z(n) < −L1 (n), where L1 > 0
EJQTDE, 2008 No. 11, p. 7

is a constant. Hence lim z(n) < −∞, a contradiction. Consequently, the later holds and
n→∞
hence lim z(n) = −∞, a contradiction to the fact that z(n + m − k) ≥ −by(n − k) which
n→∞
is bounded. Assume that ∆z(n) > 0, for n ≥ n2 . If z(n) > 0, then there exists a constant
α > 0 such that z(n) > α, for n ≥ n∗ . Thus z(n) = y(n) + p(n)y(n − m) < y(n) implies that
∆2 z(n) + G(α)q(n) < 0,
for n ≥ n3 > max{n2 , n∗ }. Hence
∞
X

Q(n) < ∞,

n=n3

a contradiction to (H1 ). Ultimately, z(n) < 0, for n ≥ n2 . In this case n→∞
lim z(n) exists. Let
it be β, 0 ≤ β < ∞. Suppose that β = 0. Due to ∆z(n) > 0, Eq.(1) becomes
1
−∆z(n) + q(n)G − z(n + m − k) < 0
b




where y(n − k) ≥ − 1b z(n + m − k). Setting − 1b z(n) = x(n), the last inequality can be written
as
∆x(n) +

1
q(n)G(x(n + m − k)) < 0
b

 

for n ≥ n3 > n2 and hence using (H7 ), we get
∆x(n) +

γ
q(n)x(n + m − k) < 0
b

 

which has no positive solution due to (H8 ), a contradiction to the fact that x(n) > 0 is a
solution. If 0 < β < ∞, then the contradiction is obivious due to (H1 ). Hence the theorem
is proved.

3

Superlinear Oscillation

This section deals with the oscillation of all solutions of Equation (1) such that G is superlinear.
Theorem 3.1
lowing conditions
(H9 )

lim
inf
n→∞

Let 0 ≤ p(n) ≤ a < ∞. Assume that (H), (H3 )-(H5 ) and the foln
P

Q(s) > 0,

s=n−k+1

where Q(n + 1) = min{q(n), q(n − m)} .

EJQTDE, 2008 No. 11, p. 8

(H10 )

∞
P

n−0

"

n+k+1
P

Q(s)ℓn

s=n+1

n+k+1
P
s=n+1

hold. Then Eq.(1) is oscillatory.

λQ(s)
(1+G(a))

!

−

n+k+1
P

q(s)ℓn

s=n+2

n+k+1
P
s=n+2

λQ(s)
(1+G(a))

!#

=∞

Proof
Proceeding as in Theorem 2.1, we get the inequality (6). Using the fact
that both z(n) and ∆z(n) are non-increasing, inequality (6) can be written as
0 > (1 + G(a))∆z(n + 1) + λQ(n + 1)G(z(n + 1 − k))
that is,
∆z(n + 1) +

λ
Q(n + 1)G(z(n + 1 − k)) < 0
(1 + G(a))

for n ≥ n3 > n2 . In view of (H10 ) and Theorem 1.2, the last inequality has no positive
solution, a contradiction. If ∆z(n) > 0 for n ≥ n2 , then proceeding as in Theorem 2.1, we
get
∞
X

Q(n + 1) < ∞.

n=n3

On the otherhand (H9 ) implies that
∞
X

Q(n + 1) = ∞,

n=0

a contradiction. Hence the theorem is proved.
Let −1 < b ≤ p(n) ≤ 0. Assume that (H1 ), (H4 ) and the follow-

Theorem 3.2
ing conditions
(H11 )

(H12 )

lim inf

n−1
P

n→∞ s=n+m−k

∞
P

n=0

"

n+k−m
P
s=n

q(s) > 0, k > m

q(s)ℓn

n+k−m
P 
s=n

−1
b



q(s) −

hold. Then every solution of (1) oscillates.
Proof

Clearly, (H11 ) implies that

!

∞
P

n+k−m
P
s=n+1

q(s)ℓn

n+k−m
P 
s=n+1

−1
b



q(s)

!#

=∞

q(n) = ∞ . Let y(n) be a non-oscillatory

n=0

solution of (1) such that y(n) > 0 for n ≥ n0 . Proceeding as in Theorem 2.1, we get the
inequality (4). Consequently, ∆2 z(n) ≤ 0 for n ≥ n1 > n0 + r shows that there are four
possible cases:

EJQTDE, 2008 No. 11, p. 9

1.

z(n) > 0, ∆z(n) < 0,

2.

z(n) < 0, ∆z(n) < 0,

3.

z(n) > 0, ∆z(n) > 0,

4.

z(n) < 0, ∆z(n) > 0,

Using the same type of reasoning as in the proof of the Theorem 2.2 we may obtain respective contradictions for the cases (1), (2) and (3). Consider the case (4). Since z(n) < 0.
Then substituting y(n − k) > ( 1b )z(n + m − k) for n ≥ n2 . Eq.(1) can be written as
1
∆x(n) − q(n)G(x(n + m − k)) < 0,
b




where x(n) = 1b z(n) > 0. In view of Theorem 1.2 and (H12 ), the last inequality has no
positive solution, a contradiction. This completes the proof of the theorem.
Theorem 3.3
Let −∞ < b ≤ p(n) < −1. Assume that all the conditions of Theorem
3.2 hold. Then every bounded solution of (1) oscillates.
Proof
Let y(n) be a bounded solution of (1) such that y(n) > 0 for n ≥ n0 . Then
proceeding as in the proof of the Theorem 3.2, we have four cases, cases (1), (2) and (3)
follows from the Theorem 2.3 and case (4) follows from the Theorem 3.2. Hence the details
are omitted. This completes the proof of the theorem.
Theorem 3.4
lowing conditions
(H13 )
(H14 )

∞
R

dx
G(x)
0
∞
P

j=0

Let −∞ < b ≤ p(n) < −1. Assume that m ≥ k + 1 and the fol-

0 for n ≥ n0 . The
case y(n) < 0 for n ≥ n0 is similar. Using the same type of reasoning as in the proof of
Theorem 3.2, we consider the four cases (1), (2), (3) and (4). For the cases (1) and (3), the
discussion is same which can be followed from the Theorem 3.2 directly. Consider the case
(2). Here Eq.(1) reducess to
1
∆z(n + 1) + q(n)G z(n + 1) = ∆z(n) < 0
b




EJQTDE, 2008 No. 11, p. 10

that is,
q(n) < −

∆z(n + 1)
G



1
z(n
b

0, it is possible to find n2 > n1 such that nj ≥ n2 implies
y(nj ) > M. Let n3 ≥ n2 + K. Then Eq.(1) yields
s−1
X

q(nj )G(y(nj − k)) = −

s−1
X

∆2 z(nj ) = −∆z(s) + ∆z(n3 ).

nj =n3

nj =n3

Thus
G(M)

∞
X

q(nj ) <

nj =n3

∞
X

q(nj )G(y(nj − k)) < ∞,

nj =n3

a contradiction to (H14 ). Hence the proof of the theorem is complete.
Remark

The prototype of G satisfying the Theorems 3.1 - 3.3 may be of the form

G(u) =
and









h

−1

u 1 + (a + ℓn2 |u|)

i

, u 6= 0

0, u = 0

EJQTDE, 2008 No. 11, p. 11

g(u) =

















a, |u| > 1
(a + ℓn2 |u|)−1, 0 < |u| ≤ 1
0, u = 0.

The author is thankful to the referee, for helpful suggestions and necessary corrections in completion of this paper.

4

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(Received November 10, 2007)

EJQTDE, 2008 No. 11, p. 12